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Ayo, membuktikan Teorema Sisa 4! Sisa pembagian polinomial  oleh  adalah .  Bukti: Misalkan polinomal  dibagi  memberikan hasil bagi  dan sisa . Polinomial  dapat dituliskan sebagai berikut.    Untuk : ...(1) Untuk :...(2) Eliminasi n pada persamaan (1) dan (2).    Jumlahkan persamaan (1) dan (2) untuk memperoleh nilai .    Substitusikan nilai  dan  ke  diperoleh:

Pertanyaan

Ayo, membuktikan Teorema Sisa 4!

Sisa pembagian polinomial undefined oleh begin mathsize 14px style left parenthesis x minus k subscript 1 right parenthesis left parenthesis x minus k subscript 2 right parenthesis end style adalah begin mathsize 14px style s left parenthesis x right parenthesis equals fraction numerator x minus k subscript 2 over denominator k subscript 1 minus k subscript 2 end fraction cross times f left parenthesis k subscript 1 right parenthesis plus fraction numerator x minus k subscript 1 over denominator k subscript 2 minus k subscript 1 end fraction cross times f left parenthesis k subscript 2 right parenthesis end style

Bukti:
Misalkan polinomal undefined dibagi undefined memberikan hasil bagi begin mathsize 14px style h left parenthesis x right parenthesis end style dan sisa begin mathsize 14px style s left parenthesis x right parenthesis equals m x plus n end style.
Polinomial undefined dapat dituliskan sebagai berikut. 
begin mathsize 14px style f left parenthesis x right parenthesis equals left parenthesis x minus k subscript 1 right parenthesis left parenthesis x minus k subscript 2 right parenthesis h left parenthesis x right parenthesis plus m x plus... end style 

Untuk begin mathsize 14px style x equals k subscript 1 end style: begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell f left parenthesis k subscript 1 right parenthesis end cell equals cell left parenthesis k subscript 1 minus k subscript 1 right parenthesis left parenthesis k subscript 1 minus space k subscript 2 right parenthesis h left parenthesis k subscript 1 right parenthesis plus m k subscript 1 plus n end cell row blank equals cell 0 cross times left parenthesis k subscript 1 minus k subscript 2 right parenthesis space X space h left parenthesis k subscript 1 right parenthesis space plus space... space plus space n end cell row blank equals cell... plus m k subscript 1 plus n end cell row blank equals cell m k subscript 1 plus... end cell end table end style...(1)

Untuk begin mathsize 14px style x equals k subscript 2 end style:begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell space f left parenthesis k subscript 2 right parenthesis end cell equals cell left parenthesis k subscript 2 minus k subscript 1 right parenthesis left parenthesis k subscript 2 minus k subscript 2 right parenthesis h left parenthesis k subscript 2 right parenthesis plus m k subscript 2 plus n end cell row blank equals cell left parenthesis k subscript 2 minus... right parenthesis cross times 0 cross times h left parenthesis... right parenthesis plus m k subscript 2 plus n end cell row blank equals cell... plus m k subscript 2 plus n end cell row blank equals cell... plus n end cell end table end style...(2)
Eliminasi n pada persamaan (1) dan (2). 

Error converting from MathML to accessible text.

begin mathsize 14px style f left parenthesis k subscript 1 right parenthesis minus... equals m left parenthesis k subscript 1 minus k subscript 2 right parenthesis end style

begin mathsize 14px style left right double arrow m equals fraction numerator... negative f left parenthesis k subscript 2 right parenthesis over denominator k subscript 1 minus k subscript 2 end fraction end style 

Jumlahkan persamaan (1) dan (2) untuk memperoleh nilai begin mathsize 14px style n end style.

Error converting from MathML to accessible text.

begin mathsize 14px style f left parenthesis k subscript 1 right parenthesis plus f left parenthesis k subscript 2 right parenthesis equals m left parenthesis k subscript 1 plus k subscript 2 right parenthesis plus... end style

begin mathsize 14px style left right double arrow 2 n equals f left parenthesis k subscript 1 right parenthesis plus... negative m left parenthesis k subscript 1 plus... right parenthesis left right double arrow 2 n equals... plus f left parenthesis k subscript 2 right parenthesis minus begin inline style fraction numerator f left parenthesis k subscript 1 right parenthesis minus f left parenthesis k subscript 2 right parenthesis over denominator k subscript 1 minus k subscript 2 end fraction end style left parenthesis k subscript 1 plus k subscript 2 right parenthesis left right double arrow 2 n equals begin inline style fraction numerator left parenthesis f left parenthesis k subscript 1 right parenthesis plus... right parenthesis left parenthesis k subscript 1 minus k subscript 2 right parenthesis minus left parenthesis... negative f left parenthesis k subscript 2 right parenthesis right parenthesis left parenthesis k subscript 1 plus k subscript 2 right parenthesis over denominator k subscript 1 minus k subscript 2 end fraction end style left right double arrow 2 n equals begin inline style fraction numerator 2 k subscript 1 f left parenthesis k subscript 2 right parenthesis minus... over denominator k subscript 1 minus k subscript 2 end fraction end style left right double arrow space space n equals begin inline style fraction numerator... negative k subscript 2 f left parenthesis k subscript 1 right parenthesis over denominator k subscript 1 minus k subscript 2 end fraction end style end style  

Substitusikan nilai begin mathsize 14px style m end style dan undefined ke undefined diperoleh:

begin mathsize 14px style s left parenthesis x right parenthesis equals begin inline style fraction numerator f left parenthesis k subscript 1 right parenthesis minus f left parenthesis k subscript 2 right parenthesis over denominator k subscript 1 minus k subscript 2 end fraction end style x plus begin inline style fraction numerator k subscript 1 f left parenthesis k subscript 2 right parenthesis minus k subscript 2 f left parenthesis k subscript 1 right parenthesis over denominator k subscript 1 minus k subscript 2 end fraction end style space space space space space space equals begin inline style fraction numerator x f left parenthesis k subscript 1 right parenthesis minus x f left parenthesis k subscript 2 right parenthesis plus... negative k subscript 2 f left parenthesis k subscript 2 right parenthesis over denominator blank end fraction end style space space space space space space equals begin inline style fraction numerator left parenthesis x minus k subscript 2 right parenthesis f left parenthesis k subscript 1 right parenthesis plus left parenthesis... negative x right parenthesis f left parenthesis k subscript 2 right parenthesis over denominator k subscript 1 minus k subscript 2 end fraction end style space space space space space space equals begin inline style fraction numerator left parenthesis x minus... right parenthesis f left parenthesis k subscript 1 right parenthesis over denominator k subscript 1 minus k subscript 2 end fraction end style plus begin inline style fraction numerator left parenthesis k subscript 1 minus x right parenthesis f left parenthesis k subscript 2 right parenthesis over denominator k subscript 1 minus k subscript 2 end fraction end style space space space space space space equals begin inline style fraction numerator x minus k subscript 2 over denominator k subscript 1 minus k subscript 2 end fraction end style cross times f left parenthesis k subscript 1 right parenthesis plus begin inline style fraction numerator x minus k subscript 1 over denominator k subscript 1 minus k subscript 2 end fraction end style cross times f left parenthesis k subscript 2 right parenthesis space space space space left parenthesis terbukti right parenthesis end style 

  1. ...undefined 

  2. ...undefined 

Pembahasan Video:

Pembahasan Soal:

Ayo, membuktikan Teorema Sisa 4!

Sisa pembagian polinomial undefined oleh begin mathsize 14px style left parenthesis x minus k subscript 1 right parenthesis left parenthesis x minus k subscript 2 right parenthesis end style adalah begin mathsize 14px style s left parenthesis x right parenthesis equals fraction numerator x minus k subscript 2 over denominator k subscript 1 minus k subscript 2 end fraction cross times f left parenthesis k subscript 1 right parenthesis plus fraction numerator x minus k subscript 1 over denominator k subscript 2 minus k subscript 1 end fraction cross times f left parenthesis k subscript 2 right parenthesis end style

Bukti:
Misalkan polinomal undefined dibagi undefined memberikan hasil bagi begin mathsize 14px style h left parenthesis x right parenthesis end style dan sisa begin mathsize 14px style s left parenthesis x right parenthesis equals m x plus n end style.
Polinomial undefined dapat dituliskan sebagai berikut. 
 begin mathsize 14px style f left parenthesis x right parenthesis equals open parentheses x minus k subscript 1 close parentheses open parentheses x minus k subscript 2 close parentheses h left parenthesis x right parenthesis plus m x plus n end style 

Untuk begin mathsize 14px style x equals k subscript 1 end style: begin mathsize 14px style f left parenthesis k subscript 1 right parenthesis equals open parentheses k subscript 1 minus k subscript 1 close parentheses left parenthesis k subscript 1 minus k subscript 2 right parenthesis h left parenthesis k subscript 1 right parenthesis plus m k subscript 1 plus n space space space space space space space equals 0 cross times left parenthesis k subscript 1 minus k subscript 2 right parenthesis cross times h left parenthesis k subscript 1 right parenthesis plus m k subscript 1 plus n space space space space space space space equals 0 plus m k subscript 1 plus n space space space space space space space equals m k subscript 1 plus n end style...(1)

Untuk begin mathsize 14px style x equals k subscript 2 end style:begin mathsize 14px style f left parenthesis k subscript 2 right parenthesis equals left parenthesis k subscript 2 minus k subscript 1 right parenthesis left parenthesis k subscript 2 minus k subscript 2 right parenthesis h left parenthesis k subscript 2 right parenthesis plus m k subscript 2 plus n space space space space space space space equals left parenthesis k subscript 2 minus k subscript 1 right parenthesis cross times 0 cross times h left parenthesis k subscript 2 right parenthesis plus m k subscript 2 plus n space space space space space space space equals 0 plus space m k subscript 2 plus n space space space space space space space equals m k subscript 2 plus n end style...(2)
Eliminasi n pada persamaan (1) dan (2). 

 Error converting from MathML to accessible text.

begin mathsize 14px style f left parenthesis k subscript 1 right parenthesis minus f left parenthesis k subscript 2 right parenthesis equals m left parenthesis k subscript 1 minus k subscript 2 right parenthesis left right double arrow m equals fraction numerator f left parenthesis k subscript 1 right parenthesis minus f left parenthesis k subscript 2 right parenthesis over denominator left parenthesis k subscript 1 minus k subscript 2 right parenthesis end fraction end style

Jumlahkan persamaan (1) dan (2) untuk memperoleh nilai begin mathsize 14px style n end style.

Error converting from MathML to accessible text.

begin mathsize 14px style f left parenthesis k subscript 1 right parenthesis plus f left parenthesis k subscript 2 right parenthesis equals m left parenthesis k subscript 1 plus k subscript 2 right parenthesis plus 2 n end style 

begin mathsize 14px style left right double arrow 2 n equals f left parenthesis k subscript 1 right parenthesis plus f left parenthesis k subscript 2 right parenthesis minus m left parenthesis k subscript 1 plus k subscript 2 right parenthesis end style

begin mathsize 14px style left right double arrow 2 n equals f left parenthesis k subscript 1 right parenthesis plus f left parenthesis k subscript 2 right parenthesis minus fraction numerator f left parenthesis k subscript 1 right parenthesis minus f left parenthesis k subscript 2 right parenthesis over denominator k subscript 1 minus k subscript 2 end fraction left parenthesis k subscript 1 plus k subscript 2 right parenthesis end style

begin mathsize 14px style left right double arrow 2 n equals fraction numerator left parenthesis f left parenthesis k subscript 1 right parenthesis plus f left parenthesis k subscript 2 right parenthesis right parenthesis left parenthesis k subscript 1 minus k subscript 2 right parenthesis minus left parenthesis f left parenthesis k subscript 1 right parenthesis minus f left parenthesis k subscript 2 right parenthesis right parenthesis left parenthesis k subscript 1 plus k subscript 2 right parenthesis over denominator k subscript 1 minus k subscript 2 end fraction left right double arrow 2 n equals fraction numerator k subscript 1 f left parenthesis k subscript 1 right parenthesis plus k subscript 1 f left parenthesis k subscript 2 right parenthesis minus k subscript 2 f left parenthesis k subscript 1 right parenthesis minus k subscript 2 f left parenthesis k subscript 2 right parenthesis minus k subscript 1 f left parenthesis k subscript 1 right parenthesis plus k subscript 1 f left parenthesis k subscript 2 right parenthesis minus k subscript 2 f left parenthesis k subscript 1 right parenthesis plus k subscript 2 f left parenthesis k subscript 2 right parenthesis over denominator k subscript 1 minus k subscript 2 end fraction end style 

begin mathsize 14px style left right double arrow 2 n equals fraction numerator 2 k subscript 1 f left parenthesis k subscript 2 right parenthesis minus 2 k subscript 2 f left parenthesis k subscript 1 right parenthesis over denominator k subscript 1 minus k subscript 2 end fraction left right double arrow n equals fraction numerator k subscript 1 f left parenthesis k subscript 2 right parenthesis minus k subscript 2 f left parenthesis k subscript 1 right parenthesis over denominator k subscript 1 minus k subscript 2 end fraction end style 

Substitusikan nilai begin mathsize 14px style m end style dan undefined ke undefined diperoleh:

begin mathsize 14px style s left parenthesis x right parenthesis equals begin inline style fraction numerator f left parenthesis k subscript 1 right parenthesis minus f left parenthesis k subscript 2 right parenthesis over denominator k subscript 1 minus k subscript 2 end fraction end style x plus begin inline style fraction numerator k subscript 1 f left parenthesis k subscript 2 right parenthesis minus k subscript 2 f left parenthesis k subscript 1 right parenthesis over denominator k subscript 1 minus k subscript 2 end fraction end style space space space space space space space equals begin inline style fraction numerator x f left parenthesis k subscript 1 right parenthesis minus x f left parenthesis k subscript 2 right parenthesis plus k subscript 1 f left parenthesis k subscript 2 right parenthesis minus k subscript 2 f left parenthesis k subscript 1 right parenthesis over denominator k subscript 1 minus k subscript 2 end fraction end style space space space space space space space equals begin inline style fraction numerator left parenthesis x minus k subscript 2 right parenthesis f left parenthesis k subscript 1 right parenthesis plus left parenthesis k subscript 1 minus x right parenthesis f left parenthesis k subscript 2 right parenthesis over denominator k subscript 1 minus k subscript 2 end fraction end style space space space space space space space equals begin inline style fraction numerator left parenthesis x minus k subscript 2 right parenthesis f left parenthesis k subscript 1 right parenthesis over denominator k subscript 1 minus k subscript 2 end fraction end style plus begin inline style fraction numerator left parenthesis k subscript 1 minus x right parenthesis f left parenthesis k subscript 2 right parenthesis over denominator k subscript 1 minus k subscript 2 end fraction end style space space space space space space space equals begin inline style fraction numerator x minus k subscript 2 over denominator k subscript 1 minus k subscript 2 end fraction end style cross times f left parenthesis k subscript 1 right parenthesis plus begin inline style fraction numerator left parenthesis x minus k subscript 1 right parenthesis over denominator k subscript 2 minus k subscript 1 end fraction end style cross times f left parenthesis k subscript 2 right parenthesis end style

(Terbukti)

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Terakhir diupdate 12 April 2021

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